Jamie lost 27 pounds in one year. on average how many pounds did jamie lose per month?

Mathematics

2 answers:

never [62]3 years ago

8 0

2.25 pounds because you divide 27 divided by 12. Which the 12 is months. Hope this helps. :^)

aleksandr82 [10.1K]3 years ago

3 0

Since there are 12 months in a year, all we need to do is divide 27 pounds by 12, and we get that he lost, on average, 2.35lbs per month.

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Please find the result !

Sliva [168]

Answer:

$\displaystyle - \frac{1}{2}$

Step-by-step explanation:

we would like to compute the following limit:

$\displaystyle \lim _{x \to 0} \left( \frac{1}{ \ln(x + \sqrt{ {x}^{2} + 1} ) } - \frac{1}{ \ln(x + 1) } \right)$

if we substitute 0 directly we would end up with:

$\displaystyle\frac{1}{0} - \frac{1}{0}$

which is an indeterminate form! therefore we need an alternate way to compute the limit to do so simplify the expression and that yields:

$\displaystyle \lim _{x \to 0} \left( \frac{ \ln(x + 1) - \ln(x + \sqrt{ {x}^{2} + 1 } }{ \ln(x + \sqrt{ {x}^{2} + 1} ) \ln(x + 1) } \right)$

now notice that after simplifying we ended up with a<em> </em><em>rational</em><em> </em>expression in that case to compute the limit we can consider using L'hopital rule which states that

$\rm \displaystyle \lim _{x \to c} \left( \frac{f(x)}{g(x)} \right) = \lim _{x \to c} \left( \frac{f'(x)}{g'(x)} \right)$

thus apply L'hopital rule which yields:

$\displaystyle \lim _{x \to 0} \left( \frac{ \dfrac{d}{dx} \ln(x + 1) - \ln(x + \sqrt{ {x}^{2} + 1 } }{ \dfrac{d}{dx} \ln(x + \sqrt{ {x}^{2} + 1} ) \ln(x + 1) } \right)$

use difference and Product derivation rule to differentiate the numerator and the denominator respectively which yields:

$\displaystyle \lim _{x \to 0} \left( \frac{ \frac{1}{x + 1} - \frac{1}{ \sqrt{x + 1} } }{ \frac{ \ln(x + 1)}{ \sqrt{ {x}^{2} + 1 } } + \frac{ \ln(x + \sqrt{x ^{2} + 1 } }{x + 1} } \right)$

simplify which yields:

$\displaystyle \lim _{x \to 0} \left( \frac{ \sqrt{ {x}^{2} + 1 } - x - 1 }{ (x + 1)\ln(x + 1 ) + \sqrt{ {x}^{2} + 1} \ln( x + \sqrt{ {x }^{2} + 1} ) } \right)$

unfortunately! it's still an indeterminate form if we substitute 0 for x therefore apply L'hopital rule once again which yields:

$\displaystyle \lim _{x \to 0} \left( \frac{ \dfrac{d}{dx} \sqrt{ {x}^{2} + 1 } - x - 1 }{ \dfrac{d}{dx} (x + 1)\ln(x + 1 ) + \sqrt{ {x}^{2} + 1} \ln( x + \sqrt{ {x }^{2} + 1} ) } \right)$

use difference and sum derivation rule to differentiate the numerator and the denominator respectively and that is yields:

$\displaystyle \lim _{x \to 0} \left( \frac{ \frac{x}{ \sqrt{ {x}^{2} + 1 } } - 1}{ \ln(x + 1) + 2 + \frac{x \ln(x + \sqrt{ {x}^{2} + 1 } ) }{ \sqrt{ {x}^{2} + 1 } } } \right)$